AP Calculus AB

AP Calculus AB Practice Test: Practice Test 8

Practice Test 8 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Region bounded by $y=\sqrt{4-x^2}$ and $y=0$ for $-2\le x\le2$ is revolved about $y=-3$ . Which setup is correct?

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Question 1

Region bounded by $y=\sqrt{4-x^2}$ and $y=0$ for $-2\le x\le2$ is revolved about $y=-3$ . Which setup is correct?

$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)^2-(3)^2\big]dx$ (correct answer)
$V=\pi\int_{-2}^{2}\big[(3)^2-(\sqrt{4-x^2}+3)^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2})^2-0^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(-3-\sqrt{4-x^2})^2-(-3-0)^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)-3\big]^2dx$

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -3. To adjust, add 3 to y, outer √(4-x^2) +3, inner 0+3=3? But since lower is 0, distance 0 - (-3)=3, upper √ +3 >3. Yes. A tempting distractor is choice B, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 2

A continuous function $H$ satisfies $H(-2)=\!-7$ and $H(0)=\!-1$ . Which is guaranteed?

There exists $c\in(-2,0)$ such that $H(c)=-4$ . (correct answer)
There exists $c\in(-2,0)$ such that $H(c)=0$ .
There exists $c\in(-2,0)$ such that $H(c)=-8$ .
There exists $c\in(0,2)$ such that $H(c)=-4$ .
No value is guaranteed because both endpoint values are negative.

Explanation: The Intermediate Value Theorem (IVT) guarantees H continuous with H(-2) = -7 and H(0) = -1 attains values between -7 and -1 in (-2,0). -4 is between them, making choice A correct. Choice C is wrong as -8 < -7, and choice B because 0 > -1. A common error is thinking same-sign endpoints prevent IVT, but it guarantees intermediates, countering choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 3

A function is $S(x)=\sqrt{\dfrac{x^2+1}{x^2-1}}$ . Which differentiation rules are required to find $S'(x)$ ?

Power rule only
Chain rule with quotient rule (correct answer)
Product rule only
Implicit differentiation only
No rules beyond sum/difference are needed

Explanation: This problem requires selecting differentiation procedures for $S(x)=\sqrt{\dfrac{x^2+1}{x^2-1}}$ . This function can be viewed as a composition where the outer function is the square root and the inner function is the quotient $\dfrac{x^2+1}{x^2-1}$ . To differentiate properly, we need the chain rule for the square root, and then the quotient rule for the fraction inside. Alternatively, this could be rewritten using properties of exponents and logarithms, but the direct approach requires both chain rule and quotient rule. When you encounter a function that's a composition involving a quotient, expect to combine the chain rule with the quotient rule.

Question 4

For $x^2+xy=\ln y$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?

$\dfrac{dy}{dx}=-\dfrac{2x+y}{x-\frac{1}{y}}$ (correct answer)
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+\frac{1}{y}}$
$\dfrac{dy}{dx}=-\dfrac{2x}{x-\frac{1}{y}}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x}$
$\dfrac{dy}{dx}=\dfrac{2x+y}{x-\frac{1}{y}}$

Explanation: This problem uses implicit differentiation for x² + x y = ln y to find dy/dx. Differentiating gives 2x + y + x dy/dx = (1/y) dy/dx, product on xy, chain on ln y. Dy/dx terms: x dy/dx - (1/y) dy/dx, grouping to (x - 1/y) dy/dx. Constants: 2x + y, so dy/dx = - (2x + y) / (x - 1/y). Choice B tempts with +1/y but fails due to sign error in denominator. Recognize in logarithmic equations with polynomial terms.

Question 5

A continuous function $h$ on $[1,8]$ has critical points $x=3,7$ . If $h(1)=6,h(3)=2,h(7)=4,h(8)=1$ , where is the absolute minimum?

At $x=3$
At $x=7$
At $x=1$
At $x=1$ or $x=8$ only
At $x=8$ (correct answer)

Explanation: The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice A, at x=3, because h(3)=2 is low but the minimum is h(8)=1 at the endpoint. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.

Question 6

If $f''(x)=-(x-5)^2$ for all $x$ , on which intervals is $f$ concave up and concave down?

Concave up on $(5,\infty)$ ; concave down on $(-\infty,5)$
Concave up on $(-\infty,5)$ ; concave down on $(5,\infty)$
Concave up on no interval; concave down on $(-\infty,\infty)$ (correct answer)
Concave up on $(-\infty,\infty)$ ; concave down on no interval
Concave up on $(-\infty,5)\cup(5,\infty)$ ; concave down on no interval

Explanation: This problem involves analyzing concavity when the second derivative is always non-positive. Given f''(x) = -(x-5)², we need to determine where f''(x) > 0 (concave up) and where f''(x) < 0 (concave down). Since (x-5)² ≥ 0 for all x, we have -(x-5)² ≤ 0 for all x. The expression equals zero only when x = 5, and is strictly negative for all x ≠ 5. Therefore, f is concave down on (-∞,5) and on (5,∞), which together form (-∞,∞) except at the single point x = 5. Students might think f is concave up somewhere because they see a squared term, forgetting about the negative sign. The key insight is that when f''(x) ≤ 0 everywhere and equals zero at only isolated points, the function is concave down on its entire domain.

Question 7

If $F(x)=\sqrt{x}=x^{1/2}$ , what is the second derivative $F''(x)$ for $x>0$ ?

$F''(x)=-rac{1}{4}x^{-3/2}$ (correct answer)
$F''(x)=rac{1}{4}x^{-3/2}$
$F''(x)=-rac{1}{2}x^{-1/2}$
$F''(x)=-rac{3}{4}x^{-5/2}$
$F''(x)=rac{1}{2}x^{-1/2}$

Explanation: Fractional powers require power rule with careful exponent reduction. The first is F'(x) = (1/2) x^{-1/2}. The second is F''(x) = - (1/4) x^{-3/2}, noting the negative sign. A common error is stopping early or sign mistakes. Domain x > 0 applies. For roots, rewrite as exponents and differentiate iteratively, tracking fractions.

Question 8

A function is $w(x)=\sin\!\left(\ln\!\left(x^3+2\right)\right)$ . What is $w'(x)$ ?

$\cos\!\left(\ln(x^3+2)\right)$
$\dfrac{3x^2\cos\!\left(\ln(x^3+2)\right)}{x^3+2}$ (correct answer)
$\dfrac{\cos\!\left(\ln(x^3+2)\right)}{x^3+2}$
$3x^2\cos\!\left(\ln(x^3+2)\right)$
$\sin\!\left(\ln(x^3+2)\right)\cdot \dfrac{3x^2}{x^3+2}$

Explanation: Sine of log with polynomial argument needs chain rule. Outer sin(u), u = ln(x^3 + 2), derivative cos(u), inner u' = 3x^2/(x^3 + 2). w'(x) = cos(ln(x^3 + 2)) * 3x^2/(x^3 + 2). Common omission: missing 3x^2 in numerator. Incorrect inner derivative as just 3x^2 without denominator. Identify trig of log polynomial and apply chain. This pattern helps in higher-degree composites.

Question 9

Let $F(x)=\int_{1}^{x} g(t)\,dt$ . If $g(x)<0$ but increasing on $(1,5)$ , which describes $F$ on $(1,5)$ ?

Decreasing and concave up (correct answer)
Decreasing and concave down
Increasing and concave up
Increasing and concave down
Constant and concave up

Explanation: This problem requires analyzing accumulation functions with negative but increasing integrands. Since $F'(x) = g(x) < 0$ on $(1,5)$ , the function $F$ is decreasing throughout this interval. For concavity, $F''(x) = g'(x) > 0$ since $g$ is increasing, making $F$ concave up on $(1,5)$ . This creates a decreasing function with decreasing rate of decrease. Choice B would be incorrect since it suggests both decreasing and concave down behavior. When the integrand is negative but increasing, the accumulation function decreases but is concave up.

Question 10

A reservoir’s volume changes at rate $V'(t)$ acre-feet/day, $t$ days after measurement begins. What does $\int_{2}^{9} V'(t)\,dt$ represent?

The reservoir volume on day 9, in acre-feet
The net change in reservoir volume from day 2 to day 9, in acre-feet (correct answer)
The average reservoir volume from day 2 to day 9, in acre-feet
The rate of volume change on day 9, in acre-feet per day
The change in rate of volume change from day 2 to day 9, in acre-feet per day

Explanation: This question involves interpreting a definite integral when the integrand is a rate of change function. Since V'(t) represents the reservoir's volume change rate in acre-feet per day, integrating this rate from t=2 to t=9 days gives the total change in reservoir volume during this time period. By the Fundamental Theorem of Calculus, ∫₂⁹ V'(t)dt = V(9) - V(2), representing the net change in the reservoir's volume between day 2 and day 9 after measurement begins. Choice C is a distractor because it mentions "average reservoir volume," but that would be (1/7)∫₂⁹ V(t)dt, not the integral of V'(t). The dimensional check confirms our answer: (acre-feet/day) × days = acre-feet, showing the integral measures net volume change.

Question 11

Given the implicit relation $\dfrac{x}{y}+y=3$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?

$-\dfrac{1}{y-\frac{x}{y}}$ (correct answer)
$-\dfrac{1}{y+\frac{x}{y}}$
$-\dfrac{y}{y-\frac{x}{y}}$
$\dfrac{1}{y-\frac{x}{y}}$
$-\dfrac{1}{1-\frac{x}{y^2}}$

Explanation: This problem requires implicit differentiation of x/y + y = 3. Differentiating x/y using the quotient rule gives (y·1 - x·dy/dx)/y², and the full equation becomes (y - x·dy/dx)/y² + dy/dx = 0. Multiplying through by y² yields y - x·dy/dx + y²·dy/dx = 0, which rearranges to dy/dx(y² - x) = -y. Therefore, dy/dx = -y/(y² - x) = -1/(y - x/y). Choice B incorrectly has a plus sign in the denominator, which would result from a sign error when factoring. The recognition strategy is to clear fractions strategically and recognize when the result can be simplified by factoring out common terms.

Question 12

A temperature change uses $\dfrac{(x-1)^2}{x-1}$ ; find $\lim\limits_{x\to 1}\dfrac{(x-1)^2}{x-1}$ .

$2$
$1$
$0$ (correct answer)
$-1$
$-2$

Explanation: This limit involves a perfect square expression that creates 0/0, requiring careful algebraic simplification. The numerator (x-1)² can be written as (x-1)(x-1). The expression becomes (x-1)(x-1)/(x-1), which simplifies to (x-1) after canceling one factor of (x-1). Substituting x = 1 gives 1 - 1 = 0. Students often incorrectly cancel both factors or fail to recognize the perfect square structure. The key strategy is to expand or factor perfect squares appropriately, cancel only one instance of common factors, and substitute carefully.

Question 13

A balloon’s radius is $r(t)$ cm. Which describes the instantaneous rate of change of radius at $t=4$ seconds?

$\displaystyle \frac{r(6)-r(2)}{4}$
The secant slope from $t=4$ to $t=5$ , $r(5)-r(4)$
$\displaystyle \frac{r(4)-r(0)}{4}$
$\displaystyle \lim_{h\to 0}\frac{r(4+h)-r(4)}{h}$ (correct answer)
The total radius at $t=4$ , $r(4)$

Explanation: The instantaneous rate of change of the radius requires the derivative at t=4, which is defined through a limit process. Choice D correctly represents this using the limit definition of the derivative. Choices A and C calculate average rates over 4-second intervals, choice B gives the average rate over the interval [4,5] (despite calling it a secant slope), and choice E gives the radius value itself, not its rate of change. Students often confuse r(5)-r(4) with the instantaneous rate at t=4, but this is the average rate over that 1-second interval. The key insight: instantaneous rates capture what's happening at a single moment through limits, while average rates describe change over an interval.

Question 14

Let $S(x)=\int_{-2}^{x}\left(\frac{1}{t^2+4}\right)dt$ . What is $S'(x)$ ?

$\displaystyle \int_{-2}^{x}\left(\frac{1}{t^2+4}\right)dt$
$\displaystyle \frac{1}{t^2+4}\big|_{t=-2}$
$\dfrac{1}{x^2+4}$ (correct answer)
$\dfrac{-2x}{(x^2+4)^2}$
$\displaystyle \int_{-2}^{x}\dfrac{-2t}{(t^2+4)^2}\,dt$

Explanation: This is a classic application of the Fundamental Theorem of Calculus Part 1, which tells us that the derivative of an integral with respect to its upper limit equals the integrand evaluated at that limit. For $S(x) = \int_{-2}^{x}\frac{1}{t^2+4}\,dt$ , we get $S'(x) = \frac{1}{x^2+4}$ by substituting x for t in the integrand. The solution requires no integration or differentiation of the integrand function itself. Choice D ( $\frac{-2x}{(x^2+4)^2}$ ) represents the derivative of $\frac{1}{x^2+4}$ , which students might choose if they mistakenly differentiate after applying FTC Part 1. Remember: FTC Part 1 is complete after evaluating the integrand at x; no further differentiation is needed.

Question 15

A tank’s inflow rate is $r(t)=2e^{t}$ ; with antiderivative $R(t)=2e^{t}$ , compute $\int_{0}^{2} r(t)\,dt$ .

$2e^{0}-2e^{2}$
$2e^{2}-2e^{0}$ (correct answer)
$2e^{2}+2e^{0}$
$2e^{2}$
$2e^{0}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to 2 of 2e^t dt equals R(2) - R(0), where R is 2e^t. Evaluate R at 2 to get 2e^2, then subtract R at 0 which is 2e^0 = 2. This yields 2e^2 - 2, the net accumulation. A tempting distractor is choice A, 2e^0 - 2e^2, which inverts the limits and negates the integral. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

Question 16

Find $\displaystyle \lim_{x\to 0}\frac{\ln(1+\sin x)-\sin x}{x^2}$ .

$\dfrac{1}{2}$
$-\dfrac{1}{2}$ (correct answer)
$-1$
$0$
$1$

Explanation: 0/0; twice L'Hôpital leads to -1/2. Taylor: ln(1+sin x) - sin x ≈ - (sin x)²/2 / x² → -1/2. Tempting: ignoring higher terms. Error: sign mistake. Strategy: series for log-trig combos.

Question 17

A continuous function $h$ satisfies $h(-5)=8$ and $h(-1)=2$ . What must exist?

A value $c\in(-5,-1)$ such that $h(c)=10$ .
A value $c\in(-5,-1)$ such that $h(c)=0$ .
A value $c\in(-5,-1)$ such that $h(c)=5$ . (correct answer)
A value $c\in(-1,0)$ such that $h(c)=5$ .
No value is guaranteed because both endpoint values are positive.

Explanation: The Intermediate Value Theorem (IVT) applies to continuous h with h(-5) = 8 and h(-1) = 2, guaranteeing values between 2 and 8. Since 2 < 5 < 8, there exists c in [-5, -1] with h(c) = 5, and as 5 ≠ 8, 5 ≠ 2, c is in (-5, -1). This works despite same-sign endpoints because 5 is between them. A common error is assuming no guarantee without opposite signs, but IVT cares about the range. Thinking positive values prevent application is incorrect. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 18

A population grows at rate $P'(t)=12t^5-5t^2$ . Find an antiderivative of $P'(t)$ .

$2t^6-\frac{5}{3}t^3+C$ (correct answer)
$2t^6-5t^3+C$
$12t^6-5t^3+C$
$2t^6-\frac{5}{3}t^3$
$60t^4-10t$

Explanation: This problem requires finding an antiderivative of the population growth rate P'(t) = 12t⁵ - 5t², demonstrating basic antiderivative reasoning. To find the antiderivative, we reverse the differentiation process using the power rule for integration. For 12t⁵, we get 12t⁶/6 = 2t⁶; for -5t², we get -5t³/3 = -5/3·t³. Combining these with the constant of integration gives 2t⁶ - (5/3)t³ + C. Choice B (2t⁶ - 5t³ + C) incorrectly integrates -5t² as -5t³ instead of -(5/3)t³, forgetting to divide by the new exponent. The key integration strategy is to carefully apply the power rule formula ∫tⁿdt = t^(n+1)/(n+1) + C, ensuring you divide by the new exponent (n+1).

Question 19

For $f(x)=\dfrac{1}{x-2}$ , which limit notation represents the behavior as $x$ approaches $2$ from the right?

$\displaystyle \lim_{x\to 2^+} f(x)=+\infty$ (correct answer)
$\displaystyle \lim_{x\to 2} f(x)=0$
$\displaystyle f(2)=+\infty$
$\displaystyle \lim_{x\to 2^-} f(x)=+\infty$
$\displaystyle \lim_{x\to 2^+} f(x)=-\infty$

Explanation: The function f(x) = 1/(x-2) has a vertical asymptote at x=2, where the denominator becomes zero. As x approaches 2 from the right (x > 2), x-2 is positive and small, so f(x) becomes a large positive number, indicating the limit is +∞. The correct notation is lim_{x→2^+} f(x) = +∞, which specifies the right-hand side and the direction to positive infinity. This is valid because the one-sided limit captures the behavior without the function being defined at x=2. A common symbolic error is writing f(2) = +∞ instead of using limit notation, as the function value at x=2 is undefined, not infinite. Another mistake is confusing the sides, such as using x→2^- which would yield -∞ for this function. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.

Question 20

A function has $f'(x)>0$ on $( -\infty,0)$ and $(2,\infty)$ , but $f'(x)<0$ on $(0,2)$ . If $f''(x)>0$ on $( -\infty,\infty)$ , what is true?

Local maximum at $x=0$ and local minimum at $x=2$ . (correct answer)
Local minimum at $x=0$ and local maximum at $x=2$ .
Local maximum at both $x=0$ and $x=2$ .
Local minimum at both $x=0$ and $x=2$ .
No extrema at $x=0$ or $x=2$ .

Explanation: This question applies the first derivative test with uniform concavity. The sign pattern shows f′>0 on (-∞,0) and (2,∞) but f′<0 on (0,2), indicating f increases, then decreases, then increases again, creating a local maximum at x=0 and local minimum at x=2. The condition f″(x)>0 everywhere confirms these are indeed extrema (not just horizontal tangents). Choice B might seem appealing but reverses the extrema types by misreading the sign changes. The systematic approach is to apply the first derivative test: f′ changing from positive to negative gives local maximum, negative to positive gives local minimum.

Question 21

On $(2,8)$ , $f'(x)<0$ for all $x$ and $f''(x)>0$ for all $x$ . Which graph could be $f$ ?

Decreasing and concave down on $(2,8)$ .
Increasing and concave up on $(2,8)$ .
Decreasing and concave up on $(2,8)$ . (correct answer)
Increasing and concave down on $(2,8)$ .
Decreasing with an inflection point in $(2,8)$ .

Explanation: This problem requires multi-representation reasoning to match derivative conditions to graph characteristics. Since f'(x) < 0 for all x in (2,8), the function f is decreasing throughout this interval. Since f''(x) > 0 for all x in (2,8), the function f is concave up throughout this interval. Therefore, f is decreasing and concave up on (2,8). Choice A incorrectly states f is concave down, which would require f'' < 0. When matching derivative signs to graphs, remember that f' < 0 means decreasing and f'' > 0 means concave up, creating a graph that falls at a decreasing rate.

Question 22

A function is $\psi(x)=\ln\!\big(\ln(x+2)\big)$ . What is $\psi'(x)$ ?

$\dfrac{1}{\ln(x+2)}$
$\dfrac{1}{(x+2)\ln(x+2)}$ (correct answer)
$\dfrac{1}{(x+2)\ln(\ln(x+2))}$
$\dfrac{1}{x+2}$
$\dfrac{\ln(x+2)}{x+2}$

Explanation: ψ(x) = ln(ln(x+2)) is log of log. Derivative: 1/ln(x+2) * 1/(x+2). Structure: Nested logs. Omission: Missing inner 1/(x+2). Matches choice B. Correct. Pattern: Double logs require double chain rule.

Question 23

If $f''(x)=\dfrac{x}{(x-2)^2}$ for $x\ne2$ , where is $f$ concave up and concave down?

Concave up on $(0,2)\cup(2,\infty)$ ; concave down on $(-\infty,0)$ (correct answer)
Concave up on $(-\infty,0)$ ; concave down on $(0,2)\cup(2,\infty)$
Concave up on $(-\infty,2)$ ; concave down on $(2,\infty)$
Concave up on $(2,\infty)$ ; concave down on $(-\infty,2)$
Concave up on $(-\infty,0)\cup(0,2)$ ; concave down on $(2,\infty)$

Explanation: This question involves analyzing concavity with a rational second derivative that has a discontinuity. Given f''(x) = x/(x-2)², we need to determine the sign of f''(x) on each interval created by x = 0 (where the numerator is zero) and x = 2 (where f'' is undefined). The denominator (x-2)² is always positive for x ≠ 2, so the sign of f''(x) depends only on the sign of x. For x < 0, f''(x) < 0 (concave down); for 0 < x < 2, f''(x) > 0 (concave up); for x > 2, f''(x) > 0 (concave up). A common error is thinking that x = 2 creates a concavity change because f'' is undefined there, but since f'' has the same sign on both sides of x = 2, concavity doesn't change. The key insight is that concavity changes only where f'' changes sign, which happens at x = 0.

Question 24

For $p(x)=\dfrac{1}{(x-3)^2}$ , which limit expression correctly describes the behavior as $x\to 3$ ?

$\displaystyle \lim_{x\to 3} p(x)=0$
$\displaystyle \lim_{x\to 3^-} p(x)=-\infty$
$\displaystyle p(3)=+\infty$
$\displaystyle \lim_{x\to 3} p(x)=+\infty$ (correct answer)
$\displaystyle \lim_{x\to 3^+} p(x)=-\infty$

Explanation: For p(x) = 1/(x-3)^2, the denominator is always positive near x=3 except at the point itself, leading to +∞ from both sides. As x approaches 3 from either side, (x-3)^2 is positive small, so p(x) becomes large positive. The two-sided limit notation lim_{x→3} p(x) = +∞ is appropriate since both one-sided limits agree. This is valid for even-powered denominators causing symmetric behavior to +∞. A common error is writing one-sided notation unnecessarily or using p(3) = +∞, but limits aren't function values. Students sometimes incorrectly predict -∞ by ignoring the squaring effect. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.

Question 25

For $r(x)=\begin{cases}\dfrac{x^2-25}{x+5},&x\ne-5\\9,&x=-5\end{cases}$ , what value should replace $9$ to remove the discontinuity at $x=-5$ ?

$-10$ (correct answer)
$0$
$9$
$-5$
$10$

Explanation: The function r(x) has a removable discontinuity at x = -5 where (x² - 25)/(x + 5) is undefined. Factoring the numerator as a difference of squares gives x² - 25 = (x + 5)(x - 5), so the expression simplifies to (x + 5)(x - 5)/(x + 5) = x - 5 for x ≠ -5. The limit as x approaches -5 is lim[x→-5] (x - 5) = -5 - 5 = -10. To remove the discontinuity, we must replace 9 with -10. Students often make sign errors when substituting negative values. Always factor completely, cancel common factors, then carefully evaluate the limit at the point of discontinuity.

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